Question

Hi,
I am trying to figure out what the limit as h approaches 0 of (1-2h)^(1/h) is. I am unfamiliar with the process I am supposed to use to solve this limit. I have just been reasoning out this limit, but I keep getting the answer, 1^(infinity)=1. My thinking when working with the exponent, (1/h); as h approaches 0, the overall exponent value will near infinity. When working with the base, (1-2h), my reasoning is that 1-(a very small number close to 0), so the overall base value is 1. That is how I get 1^infinity. I checked the answer and it is different than mine, so I am confused.

Thanks in advance for any hints or help you can provide!

Answer 1

lim (1-2h)^(1/h)
take logs
lim log = 1/h log(1-2h) = log(1-2h)/h
L'Hospital:
lim log = [-2/(1-2h)]/1 = -2
so, lim = e^-2

This makes sense, since

lim (1+ax)^(1/x) = e^ax
and x = -2

For details, visit wolframalpha.com and enter

limit (1-2x)^(1/x) as x->0

and click the Step-by-Step Solution button